Problem: The equation of a circle $C$ is $x^2+y^2-12x+6y-4 = 0$. What is its center $(h, k)$ and its radius $r$ ?
To find the equation in standard form, complete the square. $(x^2-12x) + (y^2+6y) = 4$ $(x^2-12x+36) + (y^2+6y+9) = 4 + 36 + 9$ $(x-6)^{2} + (y+3)^{2} = 49 = 7^2$ Thus, $(h, k) = (6, -3)$ and $r = 7$.